To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.
Since the propagation occurs in an area of spherical figure we will have to
Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that
The relation between intensity I and 
Here,
= Permeability constant
c = Speed of light
Rearranging for the Maximum Energy and substituting we have then,
Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,
Therefore the maximum value of the magnetic field is
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as
for first resonant frequency, we have n = 1
Inserting the values
Hz
depends t what stage in the fall it is. If it is at the peak, it is fully potential. If it is in the middle, it has both. If it is at the bottom of the fall, it is completely kinetic
Answer: 27 joules
Explanation:
Work is done when force is applied on the bench over a distance. it is measured in joules.
Workdone = force x distance
= 45 N x 0.6 metres
= 27 joules
Thus, 27 joules of work is done on the bench.
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
